Find Remainder of Array Product Division

Given an array arr[] of size n and an integer k, find the remainder when the product of all elements of the array is divided by k.

Examples: 

Input: arr[] = [4, 6, 7], k = 3
Output: 0
Explanation: 4 * 6 * 7 = 168 % 3 = 0.

Input: arr[] = [1, 6], k = 5
Output: 1
Explanation: 1 * 6 = 6 % 5 = 1.

[Naive Approach] Compute Complete Product - O(n) Time and O(1) Space

Multiply all elements of the array to get the product and then return the remainder when this product is divided by k.

Working of Approach:

• Initialize a variable product as 1.
• Traverse the array and multiply every element with product.
• After processing all elements, compute product % k.
• Return the obtained remainder.

#include <bits/stdc++.h>
using namespace std;

int remArray(vector<int> &arr, int k)
{

    // Stores the complete product.
    long long product = 1;

    // Multiply all array elements.
    for (int x : arr)
    {
        product *= x;
    }

    // Return remainder after division by k.
    return product % k;
}

int main()
{
    vector<int> arr = {1, 6};
    int k = 5;

    cout << remArray(arr, k);

    return 0;
}

import java.util.*;

class GFG {

    static int remArray(int[] arr, int k)
    {

        // Stores the complete product.
        long product = 1;

        // Multiply all array elements.
        for (int x : arr) {
            product *= x;
        }

        // Return remainder after division by k.
        return (int)(product % k);
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 6 };
        int k = 5;

        System.out.println(remArray(arr, k));
    }
}

def remArray(arr, k):
    # Stores the complete product.
    product = 1

    # Multiply all array elements.
    for x in arr:
        product *= x

    # Return remainder after division by k.
    return product % k


if __name__ == '__main__':
    arr = [1, 6]
    k = 5

    print(remArray(arr, k))

using System;

public class GFG {
    // Method to calculate the remainder of product of array
    // elements divided by k.
    public static int remArray(int[] arr, int k)
    {
        // Stores the complete product.
        long product = 1;

        // Multiply all array elements.
        foreach(int x in arr) { product *= x; }

        // Return remainder after division by k.
        return (int)(product % k);
    }

    public static void Main()
    {
        int[] arr = { 1, 6 };
        int k = 5;

        Console.WriteLine(remArray(arr, k));
    }
}

function remArray(arr, k)
{

    // Stores the complete product.
    let product = 1;

    // Multiply all array elements.
    for (let x of arr) {
        product *= x;
    }

    // Return remainder after division by k.
    return product % k;
}

// Driver Code
let arr = [ 1, 6 ];
let k = 5;
console.log(remArray(arr, k));

1

[Expected Approach] Using Modular Multiplication - O(n) Time and O(1) Space

First take a remainder or individual number like arr[i] % n. Then multiply the remainder with current result. After multiplication, again take remainder to avoid overflow. This works because of distributive properties of modular arithmetic. ( a * b) % c = ( ( a % c ) * ( b % c ) ) % c 

Working of Approach:

• Initialize the answer as 1.
• Traverse each array element one by one.
• Update the answer as (answer * currentElement) % k.
• Repeat until all elements are processed.
• Return the final modulo value.

Let us understand with an example:
Input: arr[] = [1, 6], k = 5

• Initialize res = 1.
• Multiply the first element: res = (1 × 1) % 5 = 1.
• Multiply the second element: res = (1 × 6) % 5 = 6 % 5 = 1.
• All elements are processed while keeping the product modulo 5.
• Therefore, the remainder of the product is 1.

#include <bits/stdc++.h>
using namespace std;

int remArray(vector<int> &arr, int k)
{

    int res = 1;

    // Multiply each element while taking modulo.
    for (int x : arr)
    {
        res = (res * x) % k;
    }

    // Return the final remainder.
    return res;
}

int main()
{
    vector<int> arr = {1, 6};
    int k = 5;

    cout << remArray(arr, k);

    return 0;
}

import java.util.Arrays;

public class GFG {
    public static int remArray(int[] arr, int k)
    {
        int res = 1;

        // Multiply each element while taking modulo.
        for (int x : arr) {
            res = (res * x) % k;
        }

        // Return the final remainder.
        return res;
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 6 };
        int k = 5;

        System.out.println(remArray(arr, k));
    }
}

def remArray(arr, k):
    res = 1

    # Multiply each element while taking modulo.
    for x in arr:
        res = (res * x) % k

    # Return the final remainder.
    return res


if __name__ == "__main__":
    arr = [1, 6]
    k = 5

    print(remArray(arr, k))

using System;

public class GFG {
    public static int remArray(int[] arr, int k)
    {
        int res = 1;

        // Multiply each element while taking modulo.
        foreach(int x in arr) { res = (res * x) % k; }

        // Return the final remainder.
        return res;
    }

    public static void Main()
    {
        int[] arr = { 1, 6 };
        int k = 5;

        Console.WriteLine(remArray(arr, k));
    }
}

function remArray(arr, k)
{
    let res = 1;

    // Multiply each element while taking modulo.
    for (let x of arr) {
        res = (res * x) % k;
    }

    // Return the final remainder.
    return res;
}

// Driver Code
const arr = [ 1, 6 ];
const k = 5;
console.log(remArray(arr, k));

1